Sequential vs. Simultaneous Probability I'm studying for the GRE and came across the following question:
Three dice are rolled simultaneously. What is the probability that exactly two of the dice will come up as the same number?
I'm a little stuck because I worked out the probability as follows...
$$\Pr(\text{Any})\Pr(\text{Same})\Pr(\text{Not Same}) = \frac 66\times\frac16\times\frac56=\frac5{36}$$
However, the answer scheme sets it out as follows...
"There are a total of $6^3 = 216$ total possible rolls for the three dice. First figure out the probability of getting exactly two 1’s.There are $5\times 3 = 15$ ways this could happen: 112, 113, 114, 115, 116; 121, 131, 141, 151, 161; or 211, 311, 411, 511, 611.You could repeat this list of $15$ possibilities in the obvious way for exactly two 2’s, exactly two 3’s, and so on. Thus, the total number of
favorable rolls is $6 \times 15 = 90$. Because there are $216$ possible rolls, $90$ of which are favorable, the probability of getting exactly two of the three dice to show the same number is $\frac{90}{216}=\frac{5}{12}$, choice (A)"
What I'm wondering is why the order matters if we're throwing dice simultaneously? So long as 2 are exactly the same the order doesn't matter surely?
 A: P(exactly two) = [(1/6*1/6*5/6)*3]*6 = 5/12.
We select any particular number and assume that it appears on two dice. Then we multiply it by 3 since there are three ways of choosing two dice from three given dice. Finally we multiply by 6 since there can be 6 different numbers.
A: Imagine the three dice have different colours. If we throw a red die, a blue die and a green die simultaneously, then the probability that the red and blue dice show the same number and the green is different is $\frac 16\times\frac 56=\frac 5{36}$. If we want exactly two to have the same number it could be red and blue, red and green, or blue and green. Each of these options has probability $\frac 5{36}$ so the total probability is $\frac 5{36}+\frac 5{36}+\frac 5{36}=\frac 5{12}$.
A: The order does not matter, but you can assign an order to simplify the calculations as long as you account for all orders. 
Another approach is to sidestep the order issue (albeit it is implicit with
this approach) let the sample space be $\Omega=\{1,...,6\}^3$ and
note that, in this instance, if $N$ is the number of repetitions, we must
have $N \in \{1,2,3\}$.
It is easy to see that $P[N=1] = {6 \cdot 5 \cdot 4 \over 6^3}$ and
$P[N=3] = {6 \over 6^3}$ and hence
$P[N=2] = 1-(P[N=1]+P[N=3]) = {5 \over 12}$.
